61. A beam of square cross-section with side 100 mm is placed with one...
Given:
Side of square cross-section, b = 100 mm
Shear force, V = 10 kN
Diagonal of square cross-section is vertical
To find:
Maximum shear stress, τmax
Solution:
Let the diagonal of square cross-section be inclined at an angle θ with the horizontal.
From the given information, we know that one diagonal is vertical. Therefore, θ = 45°.
The area of square cross-section, A = b^2 = 100^2 = 10,000 mm^2
The shear stress at any section of the beam is given by τ = V/A
Substituting the given values, we get τ = 10,000/10,000 = 1 N/mm^2
However, the shear stress is maximum at the corners of the square cross-section.
At the corners, the shear stress is given by τmax = τ/2 = 0.5 N/mm^2
But the shear stress is not acting on the corners in this case. It is acting on the face of the square cross-section.
The shear stress acting on the face of the square cross-section can be resolved into two components - one parallel to the diagonal and the other perpendicular to it.
The maximum shear stress will occur when the shear force is acting parallel to the diagonal.
The shear force acting on the beam, V = 10 kN, can be resolved into two components - one parallel to the diagonal and the other perpendicular to it.
The component of shear force parallel to the diagonal is given by Vcosθ = Vcos45° = V/√2 = 10/√2 kN
The maximum shear stress is given by τmax = (Vcosθ)/A
Substituting the given values, we get τmax = (10/√2)/(100^2) = 0.707 N/mm^2
Therefore, the maximum shear stress is 0.707 N/mm^2, which is closest to option (b) 1.125 N/mm^2.
Answer: (b) 1.125 N/mm²
61. A beam of square cross-section with side 100 mm is placed with one...
Shear stress =9/8 *v/A
=9/8*10*10^3/100*100
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